Why is it that an inverse function is axissymmetric to its reverse function, where y=x is the symmetry axis?

Question

Today I used Wolfram Alpha to calculate the inverse function of y=e^x. I got the expected result of y=log(x), where log is the natural logarithm. I looked at the plot of the original function and its reverse function. I noticed a dashed line, which appears to be the plot of y=x, and also noticed the original function and the reverse function were symmetrical to the dashed line. Can anybody explain why that observed symmetry exists?

Answer 1

$\exp:\mathbb R \to \mathbb R,x \mapsto e^x$ induces a one-to-one function of $\mathbb R$ to $\mathbb R ^+-\{{0}\}$, which has the inverse function $\log:R ^+-\{{0}\}\to\mathbb R $. We have :$$\forall (x,y)\in \mathbb R\times(\mathbb R ^+-\{{0}\}), y=e^x \iff x=\log y \color{red}{ (1)} $$ What Wolfram Alpha partly illustrates to you is

• $G=\{(x,e^x):x\in\mathbb R\}$, what is called in mathematics the graph of $\exp$ and of which the drawing is only an illustration.
• Similarly, $G'=\{(y,\log y):y>0\}$ is the graph of $\log$.

These are two parts of $\mathbb R ^2$, the plan. In $\mathbb R ^2=\{(x,y):x\in \mathbb R, y\in \mathbb R\}$, the symmetry with respect to the first bisector, i.e. the line with equation $y=x$, is $s:\mathbb R ^2 \to \mathbb R ^2, (x,y)\mapsto (y,x)$

Then, $s(G)=\{(e^x,x): x\in \mathbb R\}=\{(y,\log y):y>0\}$ according to $\color{red}{(1)}$. Otherwise written, "the original function and the reverse function were symmetrical to the dashed line". Note that the reasoning held here is the same for any bijection and its inverse function.

[Sorry if the words are not the familiar words in English, it's a little different in my language, French.But the math is the same.]